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3 years ago

Original price of a camera : is $599.99 and the discount is 48% how much is that all together

Mathematics

2 answers:

Elden [556K]3 years ago

7 0

Answer:

$311.9

599.99*.48= 287.9952

599.99-287.9952 = 311.9048

rounded to nearest cent is $311.90

ludmilkaskok [199]3 years ago

7 0

Answer:

311.48 after discount

Step-by-step explanation:

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A new shopping mall is considering setting up an information desk manned by one employee. Based upon information obtained from s

quester [9]

Answer:

a) $P=1-\frac{\lambda}{\mu}=1-\frac{20}{30}=0.33$ and that represent the 33%

b) $p_x =\frac{\lambda}{\mu}=\frac{20}{30}=0.66$

c) $L_s =\frac{20}{30-20}=\frac{20}{10}=2 people$

d) $L_q =\frac{20^2}{30(30-20)}=1.333 people$

e) $W_s =\frac{1}{\lambda -\mu}=\frac{1}{30-20}=0.1hours$

f) $W_q =\frac{\lambda}{\mu(\mu -\lambda)}=\frac{20}{30(30-20)}=0.0667 hours$

Step-by-step explanation:

Notation

P represent the probability that the employee is idle

$p_x$ represent the probability that the employee is busy

$L_s$ represent the average number of people receiving and waiting to receive some information

$L_q$ represent the average number of people waiting in line to get some information

$W_s$ represent the average time a person seeking information spends in the system

$W_q$ represent the expected time a person spends just waiting in line to have a question answered

This an special case of Single channel model

Single Channel Queuing Model. "That division of service channels happen in regards to number of servers that are present at each of the queues that are formed. Poisson distribution determines the number of arrivals on a per unit time basis, where mean arrival rate is denoted by λ".

Part a

Find the probability that the employee is idle

The probability on this case is given by:

In order to find the mean we can do this:

$\mu = \frac{1question}{2minutes}\frac{60minutes}{1hr}=\frac{30 question}{hr}$

And in order to find the probability we can do this:

$P=1-\frac{\lambda}{\mu}=1-\frac{20}{30}=0.33$ and that represent the 33%

Part b

Find the proportion of the time that the employee is busy

This proportion is given by:

$p_x =\frac{\lambda}{\mu}=\frac{20}{30}=0.66$

Part c

Find the average number of people receiving and waiting to receive some information

In order to find this average we can use this formula:

$L_s= \frac{\lambda}{\lambda -\mu}$

And replacing we got:

$L_s =\frac{20}{30-20}=\frac{20}{10}=2 people$

Part d

Find the average number of people waiting in line to get some information.

For the number of people wiating we can us ethe following formula"

$L_q =\frac{\lambda^2}{\mu(\mu-\lambda)}$

And replacing we got this:

$L_q =\frac{20^2}{30(30-20)}=1.333 people$

Part e

Find the average time a person seeking information spends in the system

For this average we can use the following formula:

$W_s =\frac{1}{\lambda -\mu}=\frac{1}{30-20}=0.1hours$

Part f

Find the expected time a person spends just waiting in line to have a question answered (time in the queue).

For this case the waiting time to answer a question we can use this formula:

$W_q =\frac{\lambda}{\mu(\mu -\lambda)}=\frac{20}{30(30-20)}=0.0667 hours$

6 0

3 years ago

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Archy [21]

Answer:

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3 years ago

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Does anyone know the answer for 42 ?

Trava [24]

Question 42 is B because I did the math

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4 years ago

Help me please lolll

Sergeu [11.5K]

Answer:

2.423

Step-by-step explanation:

Surface that is not water: 5.1 - 3.61 = 1.49

How many times water is bigger than not water: 3.61/1.49 = 2.423

4 0

3 years ago

The letters in the given list are written on tiles, one letter per tile. A, B, C, F, G, I, K, L, O, P, Q, T, V, W, X, Z The 16 t

Butoxors [25]

Answer:

(a) $\frac{1}{16}$

(b) $\frac{1}{8}$

(d) $\frac{15}{16}$

Step-by-step explanation:

The letters in each tile are:

A, B, C, F, G, I, K, L, O, P, Q, T, V, W, X, Z

Total Number of tiles n(S)=16

(a)The probability of drawing any one of the letters

The Probability of drawing any one of the letters $=\dfrac{1}{n(S)}=\dfrac{1}{16}$

(b)The probability of drawing either an "F" or "P" tile

P(Either an F or a P Tile) = P(drawing an F Tile )+P(drawing a P Tile)

$=\dfrac{1}{16}+\dfrac{1}{16}\\=\dfrac{2}{16}\\=\dfrac{1}{8}$

(c)The probability of drawing a vowel

The Vowel Tiles are: A,I,O

Number of Vowels =3

$\text{P(Drawing a vowel)}=\dfrac{3}{16}$

(d)The probability of not drawing a "Q" tile

$\text{P(Drawing a Q tile)}=\dfrac{1}{16}\\Therefore:\\\text{P(Not Drawing a Q tile)}=P(Q^c)=1-\dfrac{1}{16}\\=\dfrac{15}{16}$

6 0

3 years ago